Estimating the Characteristic Time of Spinning Tea

Question

Suppose I have a cup of tea and stir it so it gains some angular velocity $\omega_0$. Could you estimate the characteristic time $\tau$ it takes for the tea to stop rotating (or to lose half its mechanical energy)?

When trying to estimate, I assumed that the tea was in a very long cylindrical glass and that most of its energy was lost to its viscosity $\eta$. This got me to find the angular velocity as a decreasing exponential function in the form of $\omega (t) = \omega_0 \exp (-bt)$ and got around $60$ seconds.

However, when doing an experiment, I found the time to be much bigger at around $130$ seconds. Perhaps my experimentation was faulty, but I am wondering if there are some other factors at play here. How would you estimate the characteristic time?

Edit (My calculations for the estimate)

Let us model the tea like a solid cylinder. We can take the axis to be through the center of the cylinder. Let $r$ be the distance from the central axis and let's have $\omega_0$ be the angular velocity near the center. For the sake of simplicity, the angular velocity profile, $\omega (r)$ will be linear, or $$\omega R = \omega_0 (R - r) \implies \omega (r) = \frac{\omega_0 (R - r)}{R}.$$ A viscous force will inhibit the motion of the water from the walls of the glass. The viscous force is given from Newton's law as $F = -\eta A \frac{\text{d}v}{\text{d}x}.$ So the torque acting on the walls is $$\tau = \vec r \times \vec F = \eta (2\pi R^2 H) \left(\frac{\text{d} \omega}{\text{d}r}\right)_{r = R} = \eta 2\pi R^2 H \omega.$$ There will also be a torque by the change in angular momentum, or $\tau = \Delta \vec L/\Delta t$. We can integrate to find the infintesimal moment of inertia. Note that $\text{d}I = \rho \cdot 2\pi R\text{d}r H \cdot r^2.$ We also know the angular velocity profile $\omega (r)$, so $\text{d}L = \text{d}I \omega (r).$ This translates to $$\int_{0}^{L}\text{d}L = \int_{0}^{R} \frac{2\pi \rho H \omega_0}{R} (Rr^3 - r^4)\text{d}r ,$$ $$L = \frac{\rho 2\pi H \omega_0}{R} \left(\frac{R^5}{4} - \frac{R^5}{5}\right) = \frac{\rho \pi H \omega_0 R^4}{10}.$$ The change in $L$ can now be equated to viscous torque, or $$\frac{\text{d}L}{\text{d}t} = \tau \implies \frac{\rho \pi H R^4}{10}\frac{\text{d}\omega}{\text{d}t} = -2\pi \eta \omega R^2 H.$$Separating and integrating once again using the fact that $\int 1/x = \ln x + c$ tells us that $\omega_f$ is a decreasing exponential function, or $$\omega_f = \omega_i \exp \left(-\frac{20\eta}{\rho R^2}t\right).$$For the characteristic time, we require $\omega_f = \omega_0/2$. Then we would see that $$t = \frac{\rho R^2 \ln 2}{20\eta} \approx 60\;\mathrm{s}$$ assuming $R = 4\;\mathrm{cm}$.

Answer 1

Assuming that the fluid flow is laminar, we can get an exact solution.

The Navier-Stokes Equation in Cylindrical Coordinates for given problem:

$$\frac{\partial ω }{\partial t}=\nu \left ( \frac{\partial^2 ω}{\partial r^2}+\frac{3}{r}\frac{\partial ω}{\partial r} \right )$$

$$0\leqslant r \leqslant R$$

Here $ω=ω\left ( r,t \right )$

and $\nu$ denotes kinematic viscosity. OP uses dynamic viscosity $\eta $.

These are related:

$$\nu=\frac{\eta}{\rho}$$

Boundary condition:

$$ω\left ( R,t \right )=0$$

Initial condition:

$$ω\left ( r,0 \right )=ω_{0}$$

The rest is pure mathematics. I skip math to concentrate on physics only.

The solution:

$$ω\left ( r,t \right )=-\frac{2ω_{0}R}{r}\sum_{k=1}^{\infty}\frac{J_{1}\left ( \frac{r}{R}\mu_{k} \right )}{\mu_{k}J_{0}\left ( \mu_{k}\right )}e^{-\frac{\nu \mu_{k}^{2}}{R^{2}}t}$$

where $J_{0}\left ( x \right )$ and $J_{1}\left ( x \right )$ are the Bessel functions of the first kind and $\mu_{k}$ are zeros of $J_{1}\left ( x \right )$.

The first few zeros:

$$\mu_{1}=3.832$$ $$\mu_{2}=7.016$$ $$\mu_{2}=10.173$$

We see that after a short time the first term of the series begins to dominate. So we can write

$$ω\left ( r,t \right )\approx -\frac{2ω_{0}R}{r}\frac{J_{1}\left ( \frac{r}{R}\mu_{1} \right )}{\mu_{1}J_{0}\left ( \mu_{1}\right )}e^{-\frac{t}{\tau_{1}}}$$

where

$$\tau_{1}=\frac{R^{2}}{\nu \mu_{1}^{2}}$$

is the characteristic time constant of the problem.

OP said that he used water at room temperature.

Kinematic viscosity of water at $20$ Celsius is about $0.01\frac{cm^{2}}{s}$.

Assuming $R = 4\;\mathrm{cm}$ we get:

$$\tau_{1}=\frac{4^{2}}{0.01\cdot 3.83^{2}}=110\;\mathrm{s}$$

I think, this result is in satisfactory agreement with the experimental result.